Question

Electronic configuration of four elements A, B, C and D are given below:
(A) \( 1s^2 2s^2 2p^3 \) → N
(B) \( 1s^2 2s^2 2p^4 \) → O
(C) \( 1s^2 2s^2 2p^5 \) → F
(D) \( 1s^2 2s^2 2p^2 \) → C
Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

• A. A < D < B < C
• B. A < C < B < D
• C. A < B < C < D
• D. D < A < B < C

Hint:

Just remember "FON" (Fluorine, Oxygen, Nitrogen). These are the three most electronegative elements in the entire periodic table.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself. Across a period in the periodic table, electronegativity increases as the nuclear charge increases and the atomic size decreases.

Step 2: Detailed Explanation:
All four elements belong to the second period of the periodic table: - D: Carbon (Group 14, \( Z=6 \)) - A: Nitrogen (Group 15, \( Z=7 \)) - B: Oxygen (Group 16, \( Z=8 \)) - C: Fluorine (Group 17, \( Z=9 \))
Across the period from left to right: C < N < O < F. [Image of periodic table trend for electronegativity]

Step 3: Final Answer
Following the periodic trend, the correct order is D < A < B < C.